Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.010 mol of NaOH .

Part B

250.0 mL of a buffer solution that is 0.200 M in HCHO2 and 0.305 M in KCHO2

Part C

250.0 mL of a buffer solution that is 0.265 M in CH3CH2NH2 and 0.245 M in CH3CH2NH3Cl

Express your answers using two decimal places separated by a comma.

Answer 1

part B)

moles of HCHO2 = 250 x 0.2 / 1000 = 0.05

moles of KCHO2 = 250 x 0.305 / 1000 = 0.07625

pKa = 3.74

pH = pKa + log [salt / acid]

    = 3.74 + log [0.07625 / 0.05]

initial pH = 3.92

pH = pKa + log [salt + C / acid - C]

     = 3.74 + log [0.07625 + 0.01 / 0.05 - 0.01]

final pH = 4.07

part C)

moles of CH3CH2NH2 = 250 x 0.265 / 1000 = 0.06625

moles of CH3CH2NH3Cl = 250 x 0.245 / 1000 = 0.06125

moles of NaoH added = 0.01

pOH = pKb + log [salt / base]

      = 3.20 + log [0.06125 / 0.06625]

pOH = 3.16

inital pH = 10.83

pOH = pKb + log [salt - C / base + C]

       = 3.20 + log [0.06125 - 0.01 / 0.06625 + 0.01]

pOH = 3.03

final pH = 10.97